Finding the determinant of a matrix given the adjoint

Question

My attempt:

Knowing that $$A(AdjA) = IdetA$$ I took the determinant on both sides: $$det(A)det(AdjA) = det(det(A))$$ So, $$det(A)det(AdjA) = (det(A))^3$$ $$det(AdjA) = (det(A))^2$$ $$det(A) = (AdjA)^{0.5}$$ The adjoint was calculated to be 4

Thus, $$det(A) = 4^{0.5}=2$$

Why is the answer -2?

Answer 1

Your computation does not allow you to choose between $-2$ and $2$. But then you are not using the information you have about $A $.

Assuming that as you say $$\tag {1}A (\text {Adj}\,A)=I\,\det A, $$ the $3,3$ entry in (1) is the third row of $A $ times the third column of the adjoint; and this should equal $\det A $. So $$ \det A=1\times (-1)+ 2\times (-1)+ 1\times1=-2. $$

Answer 2

Another way to is to using cofactor formula. We have $$\det(A)=a_{31}C_{31}+a_{32}C_{32}+a_{33}C_{33} =C_{31}+2C_{32}+C_{33}.$$ From the adjoint of $A$, we know that $$C_{31}=-1, C_{32}=-1, C_{33}=1.$$ Therefore, we have $$\det(A)=C_{31}+2C_{32}+C_{33}=-1-2+1=-2.$$